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xeze [42]
3 years ago
12

On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds

?
AgBr: Ksp = 5.4 x 10-13

Ag2CO3: Ksp = 8.0 x 10-12

AgCl: Ksp = 1.8 x 10-10


Ag2CO3 < AgBr < AgCl
AgBr < Ag2CO3 < AgCl
AgBr < AgCl < Ag2CO3
AgCl < Ag2CO3 < AgBr
Chemistry
2 answers:
Viktor [21]3 years ago
8 0

Answer:

The order of solubility is AgBr <   Ag₂CO₃ < AgCl

Explanation:

The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:

Ksp = (A) (B) where A and B are the molar solubilities = s²  (for compounds with 1:1  ratio).

It follows then  that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:

Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.

Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹²  with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of  the ratio of ions 2:1 in Ag2CO3,  so the answer is not obvious. But since we know that

Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋

Ksp Ag2CO3  = 2s x s = 2 s² =  8.0 x 10-12

s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶

And for AgCl

AgCl  ⇄ Ag⁺ + Cl⁻

Ksp = s² = 1.8 x 10⁻¹⁰  ∴ s = √ 1.8 x 10⁻¹⁰   = 1.3 x 10⁻⁵

Therefore, AgCl is more soluble than Ag₂CO₃

The order of solubility is AgBr <   Ag₂CO₃ < AgCl

Dominik [7]3 years ago
3 0

Answer:

5.4 x 10^-13

Explanation:

Edmentum says it is correct

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Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

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You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

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Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

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Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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#SPJ1

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